Question

The period of oscillation of a simple pendulum is given by $ T=2\pi \sqrt{\frac{l}{g}}, $ where $ l $ is about 100 cm V and is known to have 1 mm accuracy. The period is about 2s. The time of 100 oscillations is measured by a stop watch of least count s. The percentage error in g is

• A. 0.1%
• B. 1%
• C. 0.2%
• D. 0.8%

Answer 1

Answer: (C)

$ T=2\pi \sqrt{\frac{l}{g}} $
$ \Rightarrow $ $ {{T}^{2}}=4{{\pi }^{2}}\frac{l}{g} $
$ \Rightarrow $ $ g=\frac{4{{\pi }^{2}}l}{{{T}^{2}}} $
Here % error in $ l=\frac{1\,mm}{100\,cm}\times 100=\frac{0.1}{100}\times 100=0.1$%
and %$ $ error in $ T=\frac{0.1}{2\times 100}\times 100=0.05 $%
$ \therefore $ $ % $ error in $ g=$%$\,\,\text{error}\,\text{in}\,l+2($%$\,\text{error}\,\text{in}\,\text{T}) $ $ =0.1+2\times 0.05 $ $ =0.2 $%